# A property of Hilbert sphere

Let $X$ be (Edit: a closed convex subset of ) the unit sphere $Y=\{x\in \ell^2: \|x\|=1\}$ in $\ell^2$ with the great circle (geodesic) metric. (Edit: Suppose the diameter of $X$ is less than $\pi/2$.) Is it true that every decreasing sequence of nonempty closed convex sets in $X$ has a nonempty intersection? (A set $S$ is convex in $X$ if for every $x,y\in S$ the geodesic path between $x,y$ is contained in $S$.)

(I edited my original question.)

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Please make your definition of a convex set more clearer. – Hassan Muhammad Oct 31 '11 at 16:59
@Muhammad. If $x\neq -y$, this means $x,y\in S$ implies $\cos(t)x+\sin(t)u \in S$ for $t$ between $0$ and $\cos^{-1}(x,y)$, where $u$ is the unit vector in the direction of $y-(x,y)x$, $(x,y)$ being the inner product of $x,y$. – TCL Oct 31 '11 at 18:00

No. For example, let $A_n$ be the subset of $X$ consisting of vectors that are zero in the first $n$ co-ordinates.

EDIT: this assumes that when $x$ and $y$ are antipodal, convexity of $S$ containing $x$, $y$ only requires that at least one of the great-circle paths is contained in $S$. If it requires all of them, then the $A_n$ are not convex. t.b. points out in the comments that in this case we can set $A_n$ to consist of all vectors in $X$ that are zero in the first $n$ co-ordinates and non-negative in the remainder.

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Surely $A_n$ is not convex? For instance, $A_1$ contains $(0,1,0,0,\dots)$ and $(0,-1,0,0,\dots)$ but not $(0,0,0,\dots)$. – Nate Eldredge Oct 31 '11 at 14:12
@Nate: "convex" in this question means as a subset of the unit sphere, with respect to the geodesic metric. It's not yet clear how convexity works with antipodal points though. – Chris Eagle Oct 31 '11 at 14:17
Ah, I see. Thanks. – Nate Eldredge Oct 31 '11 at 14:18
Chris: I think you can fix your example by requiring that all coordinates are non-negative. Then you avoid antipodal points. – t.b. Oct 31 '11 at 14:19
@Eagle. I edited my question. Thanks. – TCL Oct 31 '11 at 14:26